The velocity, expressed in terms of ME, RE is given as [tex]V_0 = \sqrt{98.99R_E}[/tex].
The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.
Maximum height of the rocket
The maximum height reached by the rocket can be modeled using conservation of energy as shown below;
P.Ei + K.Ei = K.Ef + P.Ef
[tex]M_EgR_E + \frac{1}{2} M_EV_0^2= \frac{1}{2} M_E(\frac{V_o}{10} )^2+ M_Eg(6R_E)\\\\\frac{1}{2} M_EV_0^2 - \frac{1}{2} M_E(\frac{V_o}{10} )^2 = M_Eg(6R_E) - M_EgR_E\\\\0.495M_EV_0^2= 5gM_ER_E\\\\0.495V_0^2= 5gR_E\\\\V_0 = \sqrt{\frac{5gR_E}{0.495} } \\\\V_0 = \sqrt{98.99R_E}[/tex]
Maximum height when it is launched vertically
P.E = K.E
mgH = ¹/₂mv²
H = (¹/₂v₀²)/g
Learn more about conservation of energy here: https://brainly.com/question/166559
